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Intro to Projectile on an Incline

An inclined p...

Question

An inclined plane is making an angle $β$ with horizontal. A projectile is projected from the bottom of the plane with a speed $u$ at an angle $α$ with horizontal then the time of flight $T$ is:

T = \frac{2 u s i n (α - β)}{g c o s β}

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T = \frac{u s i n (α - β)}{g c o s β}

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T = \frac{2 u s i n (α + β)}{g c o s β}

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T = \frac{u s i n (α + β)}{g c o s β}

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Solution

The correct option is B $T = \frac{2 u s i n (α - β)}{g c o s β}$
Velocity perpendicular to the inclined plane $u_{v} = u sin (α - β)$
$a = - g cos β$
$t = T$
$s = 0$
$0 = u sin (α - β) T - \frac{1}{2} (g cos β) T^{2}$
$\Rightarrow T = \frac{2 u sin (α - β)}{g cos β}$

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An inclined plane is making an angle β with horizontal. A projectile is projected from the bottom of the plane with a speed u at an angle α with horizontal then the time of flight T is:

The correct option is B T=2usin(α−β)gcosβVelocity perpendicular to the inclined plane uv=usin(α−β)a=−gcosβt=Ts=00=usin(α−β)T−12(gcosβ)T2⇒T=2usin(α−β)gcosβ

An inclined plane is making an angle $β$ with horizontal. A projectile is projected from the bottom of the plane with a speed $u$ at an angle $α$ with horizontal then the time of flight $T$ is:

The correct option is B $T = \frac{2 u s i n (α - β)}{g c o s β}$
Velocity perpendicular to the inclined plane $u_{v} = u sin (α - β)$
$a = - g cos β$
$t = T$
$s = 0$
$0 = u sin (α - β) T - \frac{1}{2} (g cos β) T^{2}$
$\Rightarrow T = \frac{2 u sin (α - β)}{g cos β}$